KindiakMath

Equating Ranges and Kernels

Problem 1. Let V be a vector space and T : V \to V be a linear transformation. Suppose T^2 = I, where we denote I = I_V for brevity. Prove that (T-I)(V) = \mathrm{ker}(T+I).

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Solution. Under the hypothesis that T^2 = I, we have (T+I)(T-I) = O.

We will first prove (\subseteq). Fix v \in (T-I)(V). Find w \in V such that v = (T-I)(w). Then

(T+I)(v) = (T+I)(T-I)(w) = (T^2 - I)(w) = O(w) = 0,

so that v \in \mathrm{ker}(T+I). Hence, (T-I)(V) \subseteq \mathrm{ker}(T+I).

Now, we prove (\supseteq). Fix v \in \mathrm{ker}(T+I). Then

0 = (T+I)(v) = T(v) + v \quad \Rightarrow \quad v = -T(v) = T(-v).

By repeated application of this result,

v = \frac 12v + \frac 12v = T(-\frac 12 v) - (-\frac 12 v) = (T-I)(-\frac 12 v) \in (T-I)(V).

Hence, \mathrm{ker}(T+I) \subseteq (T-I)(V) .

Combining both results, (T-I)(V) = \mathrm{ker}(T+I).

—Joel Kindiak, 29 Nov 24, 2130H

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